Forced Oscillations and Resonance: Lecture Notes

2.6 Forced Oscillations and Resonance1
Oscillator equation with external force F(t): basic case assumes Fperiodic,
mx00 +cx0+kx =F0cos ωt
Many real-life situations can be modelled with this equation, for example buildings in an earthquake.
There are three standard cases.
Case 1: Beating
Take c=0 (no damping/friction) and ω6=ω0=qk
m(driving frequency 6=natural frequency).
Already found complementary function xC(t) = c1cos ω0t+c2sin ω0t.
Particular integral: guess xP(t) = acos ωt+bsin ωt. Then
mx00
P+kxP= (−maω2+ka)cos ωt+ (−mbω2+kb)sin ωt
=F0cos ωt⇐⇒ a=F0
k−mω2=F0
m(ω2
0−ω2),b=0
=⇒xP(t) = F0
m(ω2
0−ω2)cos ωt
=⇒x(t) = xP(t) + xC(t) = F0
m(ω2
0−ω2)cos ωt+c1cos ω0t+c2sin ω0t
• Sum of distinct periodic motions.
• Larger F0=⇒more motion.
•ωclose to ω0=⇒more motion.
Suppose have initial conditions x(0) = 0=x0(0)(periodic force applied to resting spring). Quickly
obtain
x(t) = F0
m(ω2
0−ω2)cos ωt−cos ω0t=2F0
m(ω2
0−ω2)sin ω0−ω
2tsin ω0+ω 2t
using a trigonometric identity. If ω0,ωclose in value, then ω0−ωω0+ω
Amplitude beats at ω0−ω
2rad/s.
x
2 4 6 8 10
t
ω0=20 ω=18
Graphics show x00 +400x=38 cos 18twhere ω0=20. Solution x(t) = sin tsin 19t=A(t)sin 19t.
High frequency vibration sin19twith periodic amplitude A(t) = sin t.
1This is an abstract summary. Study this open-book and pay attention to the numerical examples from lectures...
x
0

Case 2: Resonance
No friction (c=0)where ω=ω0Driving frequency =natural frequency
Obvious guess xP(t) = acos ω0t+bsin ω0talready solves homogeneous ODE, so try
xP(t) = at cos ω0t+bt sin ω0t
Substituting in the ODE and solving for a,bgives
xP(t) = F0
2mω0tsin ω0t
xPis also the solution corresponding to2the initial conditions x(0) = x0(0) = 0.
• Sine wave with increasing amplitude A(t) = F0
2mω0t
• Spring tears itself apart!
Example Consider the IVP
(4x00 +64x=4 cos ωt
x(0) = x0(0) = 0
The natural frequency is ω0=√k/m=4 rad/s. The four graphs show the solutions for four
different driving frequencies: the last case is resonance.
0.005
0
x
2 4 6 8 10
t
ω=20 x(t) = 1
192 sin 8tsin 12t
0.1
0
x
2 4 6 8 10
t
ω=6x(t) = 1
10 sin tsin 5t
0.1
0
x
2 4 6 8 10
t
ω=5x(t) = 2
9sin t 2sin 9t 2
2
4
−4
−2
0
x
2 4 6 8 10
t
ω=4x(t) = 1
8tsin 4t
2It is also what we obtain by taking limω→ω0x(t)from the previous slide
2

Summary of undamped-driven motion The initial value problem mx00 +kx =F0cos ωtwith initial
conditions x(0) = 0=x0(0)has solution
x(t) = 
2F0/m
ω2
0−ω2sin ω0−ω
2tsin ω0+ω 2tif ω6=ω0
F0
2mω0tsin ω0tif ω=ω0
where ω0=√k/m. As ω→ω0low frequency beats of increasing amplitude occur.
The clickable animation is generated with the same F0,mthroughout
Case 3: Damped-driven motion (practical resonance)
c>0 mx00 +cx0+kx =F0cos ωt x00 +2px0+ω2
0x=F0
mcos ωt
Transient and steady-periodic solutions Three types:
Damping Condition Complementary Function xC(t)
Overdamping c2>4km e−pt c1e−√p2−ω2
0t+c2e√p2−ω2 0t
Critical damping c2=4km (c1+c2t)e−pt
Underdamping c2<4km e−ptc1cos ω1t+c2sin ω1twhere ω1=qω2
0−p2
•p>0=⇒xCtransient: lim
t→∞xC(t) = 0
• Particular integral: standard guess xP(t) = acos ωt+bsin ωtalways works.
• Regardless of initial conditions, x(t) = xP(t)for large t:steady-periodic solution.
Example Find the steady-periodic solution to x00 +3x0+2x=cos ωt.
Try xP(t) = acos ωt+bsin ωt. Substitute in the ODE:
(−aω2+3bω+2a)cos ωt+ (−bω2−3aω+2b)sin ωt=cos ωt
=⇒a=ω2−2
9ω2+ (ω2−2)2,b=3ω (ω2−2)2+9ω2
=⇒xP(t) = 1
(ω2−2)2+9ω2(ω2−2)cos ωt+3ωsin ωt=1 (ω2−2)2+9ω2cos (ωt−γ)
where γis the phase angle.
3
x
2 4 6 8 10
t
ω0=50 ω=48

General situation Long-term solution is
xP(t) = F0/m
(ω2
0−ω2)2+ (2pω)2(ω2
0−ω2)cos ωt+2pωsin ωt
=F0
mq(ω2
0−ω2)2+ (2pω)2cos(ωt−γ)where tan γ=2pω
ω2
0−ω2
• Amplitude is a function of frequency ω.
• Maximum amplitude when denominator minimal: a little calculus shows this is when
ω=(qω2
0−2p2if 2p2<ω2
0(requires very light damping c2<2km)
0 if 2p2≥ω2
0
In the first situation this is known as practical resonance.
The animation shows the steady periodic solution for the equation x00 +16x0+324x=cos ωt, namely
xP(t) = 1
p(182−ω2)2+ (16ω)2cos(ωt−γ)
for different values of driving frequency ω. Note that the practical resonant frequency (maximum
amplitude) occurs when ω=14.
4
x
2 4 6 8 10
t
ω0=18 qω2
0−2p2=14 ω=1

Glass smashing!
Model transverse motion of lip of a wine glass by3
1
10 x00 +1 5x0+1, 000, 000x=F0cos ωt
F0cos ωtmodels vibration of the air due to ambient sound.
Unforced motion (F0=0) of the glass is
xC(t) = e−t(c1cos ω1t+c2sin ω1t)
where
ω1
2π=ω1 2πqω2
0−p2=ω1
2πqk
m−c2
4m2=503.2920959 Hz
(≈1 octave above middle C)
0
x
1 2 3
t
Practical resonance occurs when
f=ω
2π=1 2πqω2
0−2p2=503.2920707 Hz
The animation shows the steady periodic solution for tiny variations of driven frequency fnear the
practical resonant frequency.
A singer must be very loud + accurate to crack the glass...
3Small mass, small damping, high spring constant
5
x
0.1
t
f=Practical Resonant Frequency +−5Hz